Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![### Problem Statement
**2) Determine if the following series converges or diverges using the Integral Test. Show your work carefully and detailed.**
\[ \sum_{n=1}^{\infty} \frac{e^{1/n}}{n^2} \]
### Detailed Explanation
#### Integral Test
To apply the Integral Test, we need to check if \( f(x) = \frac{e^{1/x}}{x^2} \) is continuous, positive, and decreasing for \( x \geq 1 \).
**Step 1: Verify if \( f(x) \) is continuous, positive, and decreasing**
- **Continuity**: \( e^{1/x} \) is continuous for all \( x \neq 0 \) and \( x^{-2} \) is continuous for \( x \neq 0 \). Hence, \( f(x) \) is continuous for \( x \geq 1 \).
- **Positivity**: \( e^{1/x} > 0 \) and \( x^{-2} > 0 \) for \( x \geq 1 \), thus \( f(x) > 0 \) for \( x \geq 1 \).
- **Decreasing**: To check if \( f(x) \) is decreasing, we can find the derivative \( f'(x) \) and see if it is less than or equal to 0 for \( x \geq 1 \).
**Step 2: Find \( f'(x) \)**
Let \( f(x) = \frac{e^{1/x}}{x^2} \).
Using the quotient rule for derivatives:
\[ f'(x) = \frac{d}{dx} \left( \frac{e^{1/x}}{x^2} \right ) = \frac{(e^{1/x})' \cdot x^2 - e^{1/x} \cdot (x^2)'}{(x^2)^2} \]
First, calculate \( (e^{1/x})' \):
\[ (e^{1/x})' = e^{1/x} \cdot \left( \frac{d}{dx} \left( \frac{1}{x} \right) \right) = e^{1/x} \cdot \left( -\frac{1}{](https://content.bartleby.com/qna-images/question/7dbb4ae4-0d65-4baa-9481-63f79be91eca/252af09d-b419-4849-ab35-a49c4d938b08/bthxtvf_reoriented.jpeg)
Transcribed Image Text:### Problem Statement
**2) Determine if the following series converges or diverges using the Integral Test. Show your work carefully and detailed.**
\[ \sum_{n=1}^{\infty} \frac{e^{1/n}}{n^2} \]
### Detailed Explanation
#### Integral Test
To apply the Integral Test, we need to check if \( f(x) = \frac{e^{1/x}}{x^2} \) is continuous, positive, and decreasing for \( x \geq 1 \).
**Step 1: Verify if \( f(x) \) is continuous, positive, and decreasing**
- **Continuity**: \( e^{1/x} \) is continuous for all \( x \neq 0 \) and \( x^{-2} \) is continuous for \( x \neq 0 \). Hence, \( f(x) \) is continuous for \( x \geq 1 \).
- **Positivity**: \( e^{1/x} > 0 \) and \( x^{-2} > 0 \) for \( x \geq 1 \), thus \( f(x) > 0 \) for \( x \geq 1 \).
- **Decreasing**: To check if \( f(x) \) is decreasing, we can find the derivative \( f'(x) \) and see if it is less than or equal to 0 for \( x \geq 1 \).
**Step 2: Find \( f'(x) \)**
Let \( f(x) = \frac{e^{1/x}}{x^2} \).
Using the quotient rule for derivatives:
\[ f'(x) = \frac{d}{dx} \left( \frac{e^{1/x}}{x^2} \right ) = \frac{(e^{1/x})' \cdot x^2 - e^{1/x} \cdot (x^2)'}{(x^2)^2} \]
First, calculate \( (e^{1/x})' \):
\[ (e^{1/x})' = e^{1/x} \cdot \left( \frac{d}{dx} \left( \frac{1}{x} \right) \right) = e^{1/x} \cdot \left( -\frac{1}{
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